G

G - Ordering Tasks

题目：

John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task is only possible if other tasks have already been executed.

Input

The input will consist of several instances of the problem. Each instance begins with a line containing two integers, 1 <= n <= 100 and m. n is the number of tasks (numbered from 1 to n) and m is the number of direct precedence relations between tasks. After this, there will be m lines with two integers i and j, representing the fact that task i must be executed before task j. An instance with n = m = 0 will finish the input.

Output

For each instance, print a line with n integers representing the tasks in a possible order of execution.

Sample Input

5 4

1 2

2 3

1 3

1 5

0 0

Sample Output

1 4 2 5 3

题目大意：

约翰有n个任务要做， 不幸的是，这些任务并不是独立的，执行某个任务之前要先执行完其他相关联的任务。

c++代码：

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<algorithm>using namespace std;int n,m;int one[105];int two[105];int degree[105]; int book[105];//表示他们都没有输出 int main(){   while(scanf("%d%d",&n,&m)!=EOF)   {      if(n==0&&m==0) break;      memset(degree,0,sizeof(degree));      memset(book,0,sizeof(book));      int i,j;      int x,y;      for(i=1;i<=m;i++)      {      scanf("%d%d",&x,&y);      degree[y]++;      one[i]=x;      two[i]=y;       }      int k=0;      int flag;      for(int i = 1;; i=(i+1)%n)   {   if(i==0)i=n;   flag=0;      if(degree[i]==0&&book[i]==0)   {        book[i]=1;        flag=1;        for(int j=1;j<=m;j++)        {          if(one[j]==i)    degree[two[j]]--;        }       }       if(flag)      {          k++;          if(k<n)printf("%d ",i);          else if(k==n)    {    printf("%d\n",i);     break;     }       }    }   }return 0;}